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The Quadratic Assignment Problem (QAP)

• common mathematical formulation for intra-company
  location problems
• cost of an assignment is determined by the
  distances and the material flows between all
  given entities
• each assignment decision has direct impact on the
  decision referring to all other objects

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The Quadratic Assignment Problem (QAP)

• Activity relationship charts
• graphical method for representing the
  desirability of locating pairs of
  machines/operations near to each other
• common letter codes for classification of
  closeness ratings
• A Absolutely necessary. Because two
  machines/operations use the same equipment or
  facilities, they must be located near each
  other.
• E Especially important. The facilities may for
  example require the same personnel or records.
• I Important. The activities may be located in
  sequence in the normal work flow.

Nahmias, S. Production and Operations Analysis,
4th ed., McGraw-Hill, 2000, Chapter 10
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The Quadratic Assignment Problem (QAP)

• common letter codes for classification of
  closeness ratings
• O Ordinary importance. It would be convenient to
  have the facilities near each other, but it is
  not essential.
• U Unimportant. It does not matter whether the
  facilities are located near each other or not.
• X Undesirable. Locating a welding department near
  one that uses flammable liquids would be an
  example of this category.
• In the original conception of the QAP a number
  giving the reason for each closeness rating is
  needed as well.
• In case of closeness rating X a negative value
  would be used to indicate the undesirability of
  closeness for the according machines/operations.

Nahmias, S. Production and Operations Analysis,
4th ed., McGraw-Hill, 2000, Chapter 10
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The Quadratic Assignment Problem (QAP)

• Example Met Me, Inc., is a franchised chain of
  fast-food hamburger restaurants. A new restaurant
  is being located in a growing suburban community
  near Reston, Virginia. Each restaurant has the
  following departments
• 1. Cooking burgers
• 2. Cooking fries
• 3. Packing and storing burgers
• 4. Drink dispensers
• 5. Counter servers
• 6. Drive-up server

Nahmias, S. Production and Operations Analysis,
4th ed., McGraw-Hill, 2000, Chapter 10
5
The Quadratic Assignment Problem (QAP)
Activity relationship diagram for the example
problem
Nahmias, S. Production and Operations Analysis,
4th ed., McGraw-Hill, 2000, Chapter 10
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The Quadratic Assignment Problem (QAP)

• Mathematical formulation
• we need both distances between the locations and
  material flow between organizational entities
  (OE)
• n organizational entities (OE), all of them are
  of same size and can therefore be interchanged
  with each other
• n locations, each of which can be provided withe
  each of the OE (exactly 1)
• thi ... transp. intensity, i.e. material flow
  between OE h and OE i
• djk ... distance between j and location k (not
  implicitly symmetric)
• Transportation costs are proportional to
  transported amount and distance.

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The Quadratic Assignment Problem (QAP)

• If OE h is assigned to location j and OE i to
  location k
• the transportation cost per unit transported from
  OE h to OE i is determined by djk
• we determine the total transportation cost by
  multiplying djk with the material flow between OE
  h zu OE i which is thi

? Cost thi djk
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The Quadratic Assignment Problem (QAP)

• Similar to the LAP
• binary decision variables
• If OE h ? location j (xhj 1)
• and OE i ? location k (xik 1)
• ? Transportation cost per unit transported from
  OE h to OE i
• ? Total transportation cost

xik 1
xhj 1
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The Quadratic Assignment Problem (QAP)

• Objective Minimize the total transportation
  costs between all OE

Quadratic function ? QAP
Constraints
Similar to LAP!!!
für h 1, ... , n ... each OE h assigned to
exactly 1 location j
für j 1, ... , n ... each location j is
provided with exactly 1 OE h
0 or 1 ... binary decision variable
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The Quadratic Assignment Problem (QAP)

• Example Calculate cost for 3 OE (1 ,2 ,3) and 3
  locations (A, B, C)

Distances between locations djk
Material flow thi
1 possible solution 1 ? A, 2 ? B, 3 ? C, i.e.
x1A 1, x2B 1, x3C 1, all other xij 0 All
constraints are fulfilled. Total transportation
cost 00 11 21 12 00 12 33
11 00 17
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The Quadratic Assignment Problem (QAP)

• This solution is not optimal since OE 1 and 3
  (which have a high degree of material flow) are
  assigned to locations A and C (which have the
  highest distance between them).

A better solution would be. 1 ? C, 2 ? A and 3
? B, i.e. x1C 1, x2A 1, x3B 1. with total
transportation cost 00 31 11 22 00
21 13 11 00 14
Material flow
Distances
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The Quadratic Assignment Problem (QAP)

• We resorted the matrix

such that row and columns appear the following
sequence 1 ? C, 2 ? A and 3 ? B, i.e C, A,
B(it is advisable to perform the resorting in 2
steps first rows than columns or the other way
round)
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The Quadratic Assignment Problem (QAP)

• Starting heuristics
• refer to the combination of one of the following
  possibilities to select an OE and a location.
• the core is defined by the already chosen OE
• After each iteration another OE is added to the
  core due to one of the following priorities

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Selection of (non-assigned) OE

• A1 those having the maximum sum of material flow
  to all (other) OE
• A2 a) those having the maximum material flow to
  the last-assigned OE
• b) those having the maximum material flow to an
  assigned OE
• A3 those having the maximum material flow to all
  assigned OE (core)
• A4 random choice

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Selection of (non-assigned) locations

• B1 those having the minimum total distance to all
  other locations
• B2 those being neighbouring to the last-chosen
  location
• B3 a) those leading to the minimum sum of
  transportation cost to the core
• b) like a) but furthermore we try to exchange
  the location with neighboured OE
• c) a location (empty or allocated) such that the
  sum of transportation costs within the new core
  is minimized (in case an allocated location is
  selected, the displaced OE is assigned to an
  empty location)
• B4 random choice

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Example

• Combination of A1 and B1
• Arrange all OE according to decreasing sum of
  material flow
• Arrange all locations according to increasing
  distance to all other locations
• Manhatten-distance between locations. (matrix is
  symmetric -gt consideration of the matrix triangle
  is sufficient)

A B C
D E F
G H I
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Example Material flow
OE 1 2 3 4 5 6 7 8 9 ?
1 - - - - 3 - - - -
2   - 3 1 2 - 4 - -
    - 3 5 2 - 3 4
4       - - - 1 - -
5         - 2 2 1 -
6           - - - -
7             - - -
8               - -
9                 -
3
10
20
3
5
15
4
7
4
4
Sequence of OE (according to decreasing material
flow) 3, 5, 2, 7, 4, 6, 8, 9, 1
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Example - Distances
L A B C D E F G H I ?
A - 1 2 1 2 3 2 3 4
B   - 1 2 1 2 3 2 3
C     - 3 2 1 4 3 2
D       - 1 2 1 2 3
        - 1 2 1 2
F           - 3 2 1
G             - 1 2
H               - 1
I                 -
18
15
18
15
12
E
15
18
15
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Sequence of locations (according to increasing
distances) E, B, D, F, H, A, C, G, I
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Example - Assignment

• Sequence of OE 3, 5, 2, 7, 4, 6, 8, 9, 1
• Sequence of locations E, B, D, F, H, A, C, G, I
• Assignment

OE 1 2 3 4 5 6 7 8 9
Loc. I D E H B A F C G
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Example Total costs
OE 1 2 3 4 5 6 7 8 9
1 - - - - 33 - - - -
2   - 31 12 22 - 42 - -
3     - 31 51 22 - 32 42
4       - - - 12 - -
5         - 21 22 11 -
6           - - - -
7             - - -
8               - -
9                 -
OE 1 and 5 are assigned to locations I and B ? 3
(Distance 1-5) 3 (Flow I-B)
Total cost 61
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The Quadratic Assignment Problem (QAP)

• Improvement heuristics
• Try to improve solutions by exchanging OE-pairs
  (see the introducing example)
• Check if the exchange of locations of 2 OE
  reduces costs.
• Exchange of OE-triples only if computational time
  is acceptable.
• There are a number of possibilities to determine
  OE-pairs (which should be checked for an exchange
  of locations)

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The Quadratic Assignment Problem (QAP)

• Selection of pairs for potential exchanges
• C1 all n(n - 1)/2 pairs
• C2 a subset of pairs
• C3 random choice
• Selection of pairs which finally are exchanged
• D1 that pair whose exchange of locations leads
  to the highest cost reduction. (best pair)
• D2 the first pair whose exchange of locations
  leads to a cost reduction. (first pair)

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The Quadratic Assignment Problem (QAP)

• Solution quality
• Combination of C1 and D1
• Quite high degree of computational effort.
• Relatively good solution quality
• A common method is to start with C1 and skip to
  D1 as soon as the solution is reasonably good.
• Combination of C1 and D1 is the equivalent to
  2-opt method for the TSP
• CRAFT
• Well-known (heuristic) solution method
• For problems where OE are of similar size CRAFT
  equals a combination of C1 and D1

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The Quadratic Assignment Problem (QAP)

• Random Choice (C3 and D2)
• Quite good results
• the fact that sometimes the best exchange of all
  exchanges which have been checked leads to an
  increase of costs is no disadvantage, because it
  reduces the risk to be trapped in local optima
• The basic idea and several adaptions/combinations
  of A, B, C, and D are discussed in literature

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Umlaufmethode

• Heurisitic method
• Combination of starting and improvement
  heuristics
• Components
• Initialization (i 1) Those OE having the
  maximum sum of material flow A1 is assigned to
  the centre of locations (i.e. the location having
  the minimum sum of distances to all other
  locations B1).
• Iteration i (i 2, ... , n) assign OE i

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Umlaufmethode

• Part 1 Selection of OE and of free location
• select those OE with the maximum sum of material
  flow to all OE assigned to the core A3
• assign the selected OE to a free location so that
  the sum of transportation costs to the core
  (within the core) is minimized B3a

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Umlaufmethode

• Part 2 Improvement step in iteration i 4
• check pair wise exchanges of the last-assigned OE
  with all other OE in the core C2
• if an improvement is found, the exchange is
  conducted and we start again with Part 2 D2

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Example Part 1

• Initialization (i 1)
• E centre
• Assign OE 3 to centre.

A B C
D E 3 F
G H I
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Sequence of assignment
i 1 2 3 4 5 6 7 8 9
OE 3              
1 0              
2 3              
3 ?              
4 3              
5 5              
6 2              
7 0              
8 3              
9 4              
5
7
2
4
6
8
9
1
i 9
3
0
0
0
?
0
0
0
i 3 2 highest mat.flow to core (3,5)
?
2
i 1 assign 3 first
i 5
0
1
1
?
?
i 2 5 highest mat.flow to 3
2
0
0
0
?
i 6
2
4
i 4
?
1
0
0
i 7
0
0
?
0
0
0
0
0
0
?
i 8
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Example Part 1Iteration i 2

• The maximum material flow to OE 3 is from OE 5
• Distances dBE dDE dFE dHE 1 equally
  minimal ? select D,
• In iteration i 2 OE 5 is assigned to D-5.

A B C
D 5 E 3 F
G H I
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Example Part 1Iteration i 3

• The maximum material flow to the core (3,5) is
  from OE 2
• Select location X, such that dXE?t23 dXD?t25
  dXE?3 dXD?2 is minimal (A, B, F, G od. H) X
  A dAE?3 dAD?2 2?3 1?2 8 X B dBE?3
  dBD?2 1?3 2?2 7 X F dFE?3 dFD?2
  1?3 2?2 7 X G dGE?3 dGD?2 2?3 1?2
  8 X H dHE?3 dHD?2 1?3 2?2 7
• B, F or H ? B is selected
• In iteration i 3 we assign OE to B

A B 2 C
D 5 E 3 F
G H I
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Example Part 1Iteration i 4

• The maximum material flow to the core (2,3,5) is
  from OE 7 (2, 3, 5)
• Select location X, such thath dXE?t73 dXD?t75
  dXB?t72 dXE?0 dXD?2 dXB?4 is minimal
• according to the given map -gt A is the best
  choice
• In iteration i 4 we tentatively assign OE 7 to
  location A

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Example Part 2

• Try to exchange A with E, B or D and calculate
  the according costsOriginal assignement (Part
  1)
• E-3, D-5, B-2, A-7 Cost 1?51?32?02?21?21?4
  18try E-3, D-5, A-2, B-7 Cost
  1?52?31?01?22?21?4 21try E-3, A-5, B-2,
  D-7 Cost 2?51?31?01?21?22?4
  25try A-3, D-5, B-2, E-7 Cost
  1?51?32?02?21?21?4 18
• Exchanging A with E would be possible but does
  not lead to a reduction of costs. Thus, we do not
  perform any exchange but go on with the solution
  determined in part 1and so on

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Example Part 2

• After 8 iterations without part 2 Cost 54
• With part 2 (last-assigned OE (9) is to be
  exchanged with OE 4) Cost 51
• While a manual calculation of larger problems is
  obviously quite time consuming an implementation
  and therefore computerized calculation is
  relatively simple